Chapter 13 of the NCERT Class 10 Mathematics textbook is Statistics (English edition), printed from page 170 to page 201. It takes the mean, median and mode you learned in Class 9 and reworks them for grouped data — data already condensed into class intervals.
The official NCERT PDF is right below, and the rest of this page explains what the chapter teaches so you can revise without opening a second tab.
Download the NCERT Class 10 Maths Statistics Chapter 13 PDF from the official NCERT website and keep it open while you study — the chapter’s worked examples and frequency tables are the working text behind every formula explained on this page.
Chapter 13 Statistics at a Glance
Before the formulas, here is the map of the whole chapter. NCERT organises Chapter 13 into five sections: 13.1 Introduction, 13.2 Mean of Grouped Data, 13.3 Mode of Grouped Data, 13.4 Median of Grouped Data and 13.5 Summary. Each teaching section ends with one exercise: Exercise 13.1 after the mean, Exercise 13.2 after the mode, Exercise 13.3 after the median.
The count table on this page lists how much the chapter holds — its sections, tables, formulas and exercise questions — all as printed in the book. Use it as a revision checklist: if you have attempted at least one question of each type, you have covered the chapter.
What This Chapter Covers: From Class 9 Measures to Grouped Data
The chapter extends what you already know. In Class 9 you studied ungrouped and grouped frequency distributions, bar graphs, histograms and frequency polygons, and the mean, median and mode of ungrouped data (NCERT, p. 170). Chapter 13 reworks those three measures for data that has been condensed into class intervals.
The new idea is the class mark. Once values are grouped, the individual observations are lost — a student with 41 marks and one with 54 both fall inside the class 40–55. The chapter assumes the frequency of each class is centred at its midpoint, so the class mark represents every observation in that class (NCERT, p. 173).
Every mean formula in this chapter rests on that assumption.
Chapter 13 also introduces cumulative frequency: the running total of frequencies up to each class. The introduction names ogives (cumulative frequency curves), but the body of this edition does not teach drawing them, so treat that mention as background (NCERT, p. 170).
Mean of Grouped Data: Direct, Assumed Mean and Step Deviation
The mean of grouped data can be found three ways. They look different but always give the same number, so the real skill is choosing the quickest one for the data in front of you.
The three formulas, with every symbol named
In all three, \(x_i\) is the class mark of the \(i\)-th class and \(f_i\) is its frequency. The Greek capital sigma, \(\sum\), means “add up”.
- Direct method (NCERT, p. 172): \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\). Multiply each class mark by its frequency, add the products, divide by the total frequency.
- Assumed mean method (NCERT, p. 176): \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\), where \(d_i = x_i – a\) and \(a\) is a chosen class mark, usually near the centre.
- Step deviation method (NCERT, p. 177): \(\bar{x} = a + h\left(\frac{\sum f_i u_i}{\sum f_i}\right)\), where \(u_i = \frac{x_i – a}{h}\) and \(h\) is the class size — or any suitable common divisor of the \(d_i\)’s.
Why they agree: subtracting \(a\) from every class mark lowers each \(f_i x_i\) by \(a f_i\), so it lowers the mean by exactly \(a\). Dividing by \(h\) scales the mean down by \(h\).
Adding \(a\) back and multiplying by \(h\) at the end undoes both changes, so the answer is the same as the direct method’s (NCERT, p. 178).
Worked example: daily wages of 30 workers (step deviation)
The daily wages of 30 workers, in rupees, fall in the classes below. Find the mean daily wage.
Classes 200–250 (4 workers), 250–300 (6), 300–350 (8), 350–400 (5), 400–450 (4), 450–500 (3).
Step 1: Write the class marks \(x_i\).
Each is the average of its class limits:
\[ x_i = 225,\ 275,\ 325,\ 375,\ 425,\ 475 \]
Step 2: Take the assumed mean \(a = 375\) and class size \(h = 50\).
With \(u_i = \frac{x_i – 375}{50}\):
\[ u_i = -3,\ -2,\ -1,\ 0,\ 1,\ 2 \]
Step 3: Multiply each \(u_i\) by its frequency \(f_i\) and sum:
\[ \sum f_i u_i = 4(-3) + 6(-2) + 8(-1) + 5(0) + 4(1) + 3(2) = -22 \]
Step 4: Apply the step-deviation formula \(\bar{x} = a + h\left( \frac{\sum f_i u_i}{\sum f_i} \right)\):
\[ \bar{x} = 375 + 50\left( \frac{-22}{30} \right) = 375 – 36.67 = 338.33 \]
Final answer: the mean daily wage is about Rs 338.33.
Which method should you use?
The chapter’s own remark on page 180 settles the choice: use the direct method when the values of \(x_i\) and \(f_i\) are small; use the assumed mean or step deviation method when they are numerically large.
| Method | Formula | When to choose it |
|---|---|---|
| Direct | \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\) | When \(x_i\) and \(f_i\) are small — fewest steps (NCERT, p. 180) |
| Assumed mean | \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\) | When \(x_i\) and \(f_i\) are numerically large |
| Step deviation | \(\bar{x} = a + h\frac{\sum f_i u_i}{\sum f_i}\) | Large numbers whose deviations share a common factor; also works for unequal class sizes by taking \(h\) as a suitable common divisor of the \(d_i\)’s (NCERT, p. 180) |
Step deviation with unequal class sizes
The chapter’s Example 3 uses cricket wicket data with classes of different widths, such as 20–60 and 150–250, and still applies the step deviation method by taking \(h = 20\) as a common divisor of the deviations (NCERT, p. 180). Here is the same idea on fresh data.
The weekly homework time of 45 students falls in classes 0–30 (5 students), 30–60 (8), 60–90 (14), 90–120 (10) and 120–180 (8).
Step 1: The class marks of 0–30, 30–60, 60–90, 90–120, 120–180 are \(15,\ 45,\ 75,\ 105,\ 150\).
Choose \(a = 75\), the middle mark.
Step 2: The deviations \(d_i = x_i – 75\) are \(-60,\ -30,\ 0,\ 30,\ 75\).
A common factor is 15, so take \(h = 15\):
\[ u_i = -4,\ -2,\ 0,\ 2,\ 5 \]
Step 3: Sum \(f_i u_i\):
\[ \sum f_i u_i = 5(-4) + 8(-2) + 14(0) + 10(2) + 8(5) = 24 \]
Step 4: Compute the mean:
\[ \bar{x} = 75 + 15\left( \frac{24}{45} \right) = 75 + 8 = 83 \]
Final answer: the mean weekly homework time is 83 minutes. Notice that \(h = 15\) is not the width of any class; it is a common divisor of the deviations, exactly as the chapter’s remark permits (NCERT, p. 180).
Why the grouped mean (62) differs from the exact mean (59.3)
The chapter’s first example does the same data twice. With the 30 individual marks, the exact mean is 59.3. Grouped into classes of width 15 and replaced by class marks, the same data gives 62 (NCERT, p. 174).
The difference is the midpoint assumption: a student who actually scored 40 is treated as scoring 47.5, one with 56 as 62.5. The grouped mean is therefore an estimate, not the exact mean. The finer the classes, the smaller the rounding this assumption causes.
Mode of Grouped Data: The Value Inside the Modal Class
The mode is the value that occurs most often. In grouped data the exact repeating value is hidden inside a class, so you first locate the class with the highest frequency — the modal class — and then interpolate inside it (NCERT, p. 184).
\[ \text{Mode} = l + \left( \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \right) \times h \]
where \(l\) is the lower limit of the modal class, \(h\) the class size, \(f_1\) the frequency of the modal class, \(f_0\) the frequency of the class immediately before it, and \(f_2\) the frequency of the class immediately after it (NCERT, p. 184).
Why the formula works: the modal class only tells you where the mode lives. The formula positions the mode inside that class by comparing the two neighbouring frequencies.
If they are equal, the mode lands at the class midpoint; a more crowded class before pulls it toward the lower limit, and a more crowded class after pushes it toward the upper limit.
Worked example: books borrowed per day
A library records the number of books borrowed per day over 40 days: 0–2 (5 days), 2–4 (8), 4–6 (12), 6–8 (9), 8–10 (4), 10–12 (2). Find the modal number of books borrowed.
Step 1: The highest frequency is 12, in the class 4–6.
So the modal class is 4–6.
- Step 1: Read off \(l = 4\), \(f_1 = 12\), \(f_0 = 8\) (class 2–4), \(f_2 = 9\) (class 6–8), and \(h = 2\).
- Step 2: Substitute into the mode formula:
\[ \text{Mode} = 4 + \left( \frac{12 – 8}{2(12) – 8 – 9} \right) \times 2 = 4 + \frac{4}{7} \times 2 = 4 + 1.14 = 5.14 \]
Final answer: the modal number of books borrowed per day is about 5.14. It sits just above the midpoint 5 because the class after the modal class (frequency 9) is more crowded than the class before it (frequency 8).
Three restrictions the chapter states about the mode:
- The chapter handles data with a single mode only; if two or more classes share the maximum frequency, the data is multimodal and is not treated here (NCERT, p. 184).
- The mode may be less than, equal to, or greater than the mean. In Example 6 the mean is 62 and the mode is 52: the most common score is below the average score (NCERT, p. 185).
- The mode of grouped data with unequal class sizes can be calculated, but the chapter does not discuss it (NCERT, p. 186).
Median of Grouped Data: Cumulative Frequency and the Median Class
The median is the middle observation. With grouped data you cannot see the middle observation, so the chapter builds a cumulative frequency column and locates the class that contains the middle of the data (NCERT, p. 189).
The cumulative frequency of a class is its frequency plus the frequencies of all classes before it. Adding the frequencies in order produces the cumulative frequency column (NCERT, pp. 190–191). The same column can be written as a less-than distribution (using upper limits) or a more-than distribution (using lower limits) (NCERT, pp. 192–193).
Finding the median class: compute \(n/2\). The median class is the class whose cumulative frequency is greater than, and nearest to, \(n/2\) (NCERT, p. 193). Then use:
\[ \text{Median} = l + \left( \frac{\frac{n}{2} – cf}{f} \right) \times h \]
where \(l\) is the lower limit of the median class, \(n\) the total number of observations, \(cf\) the cumulative frequency of the class preceding the median class, \(f\) the frequency of the median class and \(h\) the class size (NCERT, p. 193).
The most common error here is using the median class’s own cumulative frequency as \(cf\). The \(cf\) in the formula must come from the row just above the median class — it must be smaller than \(n/2\).
Worked example: marks of 40 students
Marks of 40 students fall in classes 10–15 (3 students), 15–20 (7), 20–25 (12), 25–30 (9), 30–35 (6) and 35–40 (3). Find the median mark.
Step 1: Add a cumulative frequency column:
\[ cf = 3,\ 10,\ 22,\ 31,\ 37,\ 40 \]
Step 2: \(n = 40\), so \(\frac{n}{2} = 20\).
The first cumulative frequency that reaches or passes 20 is 22, in the class 20–25.
That is the median class.
- Step 1: Read off \(l = 20\), \(cf = 10\) (cumulative frequency of the class before 20–25), \(f = 12\), \(h = 5\).
- Step 2: Apply the median formula:
\[ \text{Median} = 20 + \left( \frac{20 – 10}{12} \right) \times 5 = 20 + 4.17 = 24.17 \]
Final answer: the median mark is about 24.17 — roughly half the students scored below it and half above it.
Two question shapes the chapter uses
Less-than table first (Example 7, NCERT, p. 195). If the data is given as cumulative frequencies (“less than 140”, “less than 145”, …), rebuild the class frequencies by subtracting successive cumulative frequencies: the frequency of 140–145 is \(11 – 4 = 7\), the frequency of 145–150 is \(29 – 11 = 18\), and so on. The heights of 51 girls in that example give a median height of 149.03 cm.
Missing frequencies (Example 8, NCERT, pp. 196–197). When the median is given, write the cumulative frequencies in terms of the unknowns. The total frequency gives one equation; substituting into the median formula gives another. In Example 8, median 525 with \(n = 100\) yields \(x + y = 24\), then the median formula gives \(x = 9\), so \(y = 15\).
Mean, Median or Mode: Choosing the Right Measure
All three measures summarise a data set, but they answer different questions. The chapter closes by explaining when each one is the honest choice (NCERT, pp. 197–198).
| Measure | What it tells you | When the chapter recommends it |
|---|---|---|
| Mean | The average; uses every observation; allows comparison of whole distributions | Data with no extreme values — e.g. comparing school results (NCERT, p. 197) |
| Median | The middle, “typical” value of the data | Extreme values present — e.g. average wage, productivity rate (NCERT, p. 198) |
| Mode | The most frequent or most popular value | Finding what is most common or most in demand — e.g. a popular television programme, a colour of vehicle (NCERT, p. 198) |
Empirical relationship: \(3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean}\) (NCERT, p. 198). Using the chapter’s own example values, mean 62 and mode 52 give \(3 \times \text{Median} = 52 + 124 = 176\), so the median is about 58.7. This is an empirical approximation, useful as a quick check — not an exact law.
Reading the Chapter’s Working Tables
This chapter has no diagrams. Its visual content is the frequency tables, and statistics answers are judged on the table as much as on the final number. Three tables carry the main methods.
Table 13.3 (NCERT, p. 174) — the direct-method layout. Four columns: class interval, frequency \(f_i\), class mark \(x_i\), and product \(f_i x_i\). Each row handles one class; the last column converts “how many students” into “how many marks this class contributes”.
The bottom row holds the two totals \(\sum f_i\) and \(\sum f_i x_i\) — the only numbers the formula needs.
Table 13.11 (NCERT, p. 190) — building cumulative frequency. Each new entry is the previous total plus the next frequency: \(26 = 6 + 20\), \(50 = 26 + 24\), and so on. The final entry must equal \(n\); that equality is your built-in check that no row was skipped.
Table 13.15 (NCERT, p. 193) — spotting the median class. Frequency and cumulative frequency sit side by side. Read down the \(cf\) column until it first reaches or passes \(n/2\); that row is the median class. In the chapter’s 53-student table, \(cf\) jumps from 22 to 29 across \(n/2 = 26.5\), marking the class 60–70.
Copy these layouts in your own answer. The columns do the checking for you, and a wrong entry shows up in the running totals before it reaches the formula.
Statistics Mistakes That Cost Marks (and How to Fix Them)
Every formula in this chapter has one characteristic trap, and NCERT flags each of them at a specific page. The table below names the mistake, the correct rule, and a quick way to check yourself.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Putting a value equal to a class’s upper limit into that class | The upper limit belongs to the next class: 40 goes into 40–55, not 25–40 (NCERT, p. 173) | Re-tally any observation that sits exactly on a class boundary |
| Quoting the grouped mean as the exact mean | The class-mark assumption makes it approximate: 59.3 is exact, 62 is the grouped estimate (NCERT, p. 174) | If the raw values are available, compute the ungrouped mean and compare |
| Using any near class as \(f_0\) or \(f_2\) in the mode formula | \(f_0\) is the class immediately before the modal class, \(f_2\) immediately after (NCERT, p. 184) | Write the modal class first, then label its two neighbours before substituting |
| Using the median class’s own cumulative frequency as \(cf\) | \(cf\) is the cumulative frequency of the class preceding the median class (NCERT, p. 193) | Your \(cf\) must be smaller than \(n/2\); if it is not, you read the wrong row |
| Applying the mode or median formula to non-continuous classes | Convert to continuous classes first, e.g. 117.5–126.5, 126.5–135.5 (NCERT, p. 201) | Check that each class’s upper limit equals the next class’s lower limit; adjust by half the gap if not |
| Using the direct method on huge numbers | Direct method for small values; assumed mean or step deviation for large ones (NCERT, p. 180) | Skim the class marks first; if products such as \(92.5 \times 400\) appear, switch method |
Which Exercise Trains Which Skill: An Exam-Focused Map
Each exercise targets one measure. If you are revising tonight, this map tells you which questions to attempt for the skill you are weakest in.
| Exercise | Skill it trains | Questions with a twist |
|---|---|---|
| Exercise 13.1 (mean) | Finding the mean of grouped data by a suitable method | Q1 and Q5 ask “which method did you use, and why”; Q3 gives the mean and asks for a missing frequency; Q7 uses decimal class limits |
| Exercise 13.2 (mode) | Finding the mode, and comparing it with the mean | Q1 and Q4 ask you to “compare and interpret” the two measures; Q4 has two classes with zero frequency, so those rows contribute nothing to the mean |
| Exercise 13.3 (median) | Finding the median, including less-than tables and missing frequencies | Q1 asks median, mean and mode together; Q2 is a missing-frequency pair (median 28.5); Q3 is a less-than table; Q4 needs conversion to continuous classes; Q6 asks all three measures |
Two phrasings repeat across the exercises, and both demand a sentence of interpretation, not just a number: “compare and interpret” and “which method did you use, and why”. Practise writing one-line answers to both so a final number is never left uninterpreted.
The Chapter’s Checklist: Five Formulas to Remember
These are the formulas the chapter’s own summary asks you to carry away (NCERT, pp. 200–201). Write each from memory, with its symbols named, before you close the book.
| Measure | Formula | Symbols you must name |
|---|---|---|
| Mean — direct | \(\bar{x} = \frac{\sum f_i x_i}{\sum f_i}\) | \(x_i\) = class mark, \(f_i\) = frequency (NCERT, p. 200) |
| Mean — assumed mean | \(\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}\) | \(d_i = x_i – a\), \(a\) = assumed mean (NCERT, p. 200) |
| Mean — step deviation | \(\bar{x} = a + h\left( \frac{\sum f_i u_i}{\sum f_i} \right)\) | \(u_i = \frac{x_i – a}{h}\), \(h\) = class size or common divisor (NCERT, p. 201) |
| Mode | \(\text{Mode} = l + \frac{f_1 – f_0}{2f_1 – f_0 – f_2} \times h\) | \(l\) = lower limit of modal class, \(f_1\) its frequency, \(f_0\) before it, \(f_2\) after it (NCERT, p. 201) |
| Median | \(\text{Median} = l + \frac{\frac{n}{2} – cf}{f} \times h\) | \(l\) = lower limit of median class, \(cf\) = cumulative frequency of the class before it, \(f\) = frequency of median class (NCERT, p. 201) |
Two non-formula facts complete the checklist. Cumulative frequency of a class is the sum of its frequency and all preceding frequencies (NCERT, p. 200). And the three measures are linked by the empirical relationship \(3 \times \text{Median} = \text{Mode} + 2 \times \text{Mean}\) (NCERT, p. 198).
Related Chapters and Resources
This listing is maintained for the 2026-27 academic session using the NCERT textbook information available to us. NCERT remains the authority for confirming the latest edition.
Chapter 13 sits between Surface Areas and Volumes (Chapter 12) and Probability (Chapter 14). The Class 10 Mathematics notes hub lists every chapter of this book, the Class 10 hub covers the other subjects, and the full CBSE notes library holds every class.
For the book’s complete contents, browse the official NCERT textbook page for Class 10 Mathematics, and keep the official chapter PDF from the top of this page open while you practise.
Sources and Data Verification
This page describes the NCERT Class 10 Mathematics textbook, Chapter 13 Statistics (English edition), printed pages 170–201. It covers this chapter only, not other chapters or other editions of the book.
NCERT publishes the textbook and its official PDF; CBSE sets the examination syllabus. Textbook contents and the examinable syllabus are not always identical, so check the current official CBSE syllabus before planning what to study.
This page is maintained for the current academic session and is verified against the NCERT text cited above; no review by NCERT or CBSE is claimed.
| What the chapter holds | Count | Where it is used |
|---|---|---|
| Printed pages | 31 | |
| Sections in the chapter | 5 | |
| Tables | 47 | |
| Worked examples | 1 | solved step by step in our NCERT Solutions |
| Exercise questions | 22 | answered in our NCERT Solutions |
| Activities | 1 | |
| Official NCERT PDF | Download the chapter PDF | the chapter exactly as NCERT publishes it |
Chapter-wise notes for this subject are not published yet. The unit-wise syllabus on this page is complete and verified against the official CBSE curriculum document.
Frequently Asked Questions About This Chapter
Why do the direct, assumed mean and step deviation methods give the same mean?
Because they are the same calculation, shifted and scaled. Subtracting \(a\) from every class mark lowers the mean by exactly \(a\); dividing by \(h\) divides the mean by \(h\). Adding \(a\) back and multiplying by \(h\) restores the original mean, so all three formulas must agree (NCERT, p. 178).
What is the difference between the modal class and the mode?
The modal class is the class interval with the highest frequency. The mode is a single value inside that class, found by the mode formula. In the chapter’s Example 5, the modal class is 3–5 with frequency 8, and the mode inside it is 3.286 (NCERT, pp. 184–185).
How do I find the median class in grouped data?
Compute \(n/2\), then read down the cumulative frequency column. The median class is the first class whose cumulative frequency is greater than, and nearest to, \(n/2\). For the chapter’s 53-student data, \(n/2 = 26.5\) and the class 60–70, with cumulative frequency 29, is the median class (NCERT, p. 193).
Why is the grouped mean different from the exact mean?
Because every observation in a class is replaced by the class mark. The exact mean of the chapter’s marks data is 59.3; grouping the same data and using class marks gives 62 (NCERT, p. 174). The difference comes from the midpoint assumption, which is why the grouped mean is described as approximate.
Does Class 10 Statistics require drawing ogives?
The introduction names ogives, but the teaching sections of this edition — 13.2 to 13.4 — do not develop them, and the closing note only adds that the same continuity condition applies to ogives (NCERT, p. 201). Check the current official syllabus, because textbook contents and the examinable syllabus are not always identical.
How do I find a missing frequency when the mean or median is given?
For a missing frequency in a mean question, write \(\sum f_i x_i\) in terms of the unknown and set it equal to the given mean times the total frequency; Exercise 13.1 Q3 (mean 18) does exactly this.
For a median question, build the cumulative frequencies with the unknowns and substitute into the median formula — Example 8 shows the method: \(x + y = 24\), then \(x = 9\), so \(y = 15\) (NCERT, pp. 196–197).
Reference: NCERT Class 10 Mathematics textbook, chapter 13, official edition on ncert.nic.in.
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